The “power” of the corners: full analysis 6x6 2012-Apr-14 (this post is dedicated to beaker, jomapil, … , and all those calcudokers who like the “addition rule” so that they may start liking the “multiplication rule” as well).

When we initially deal with a puzzle like this (6x6 difficult, Apr 14, 2012, Puzzle id: 446084):

we are inclined to think of a great complexity that in fact does not exist as we will see inmediately. The puzzle itself (with an assigned solving rate of 48.7) is not “officially” considered “very” difficult (and in fact it was solved by many puzzlers). But, in my opinion, it is a very interesting calcudoku for two reasons: 1) it can be solved applying only “analytical means” and 2) because of the possibility of discussing the relevance of the value of the corners (useful in some specific puzzles) and incidentally the content of the “inner area”, etc., apart of practicing with the general “multiplication rule”.

Perhaps we could have started breaking into prime factors all products involved (knowing that the products in the border cann’t repeat a number more than twice). Though we will not need it this time (because, along the process, we will not use at any moment any sum or the “addition” rule), I have also included (in parentheses) the “addition value” for every combination:

In a quick look we see that, since in c3 or c4 we can place (at most) a single 5 and there is another 5 in the cage “540x”, to complete the necessary three 5’s in the three leftmost columns we need a third 5 inside “16+” but, in these conditions, to complete the cage “16+”, the rest is 11, we need 5 + 6 (unique) so “16+” = [5,5,6]. This has two consequences: a6 = 5 (the only place for the 5 of “540x”) and d4 = 5 (it must be a 5 in the cage “80x”).

Comment: Supposing that the puzzler does not take notice of those 5’s at a first glance there is always a method to quickly advance to the solution: applying the multiplication property (though in this case managing huge products). The multiplication of the numbers of any line in a 6x6 is 6 x 5 x 4 x 3 x 2 x 1 = 720 (named 6!, factorial of 6). The multiplication of the three numbers inside the cage “16+” by the three numbers inside the cage “2:” must be equal to:

[(720) ^ 6] / (96 x 900 x 12 x 80 x 540 x 4 x 288) = 2700,

that is, the multiplication of all the numbers in the full grid (720 x 720 x … x 720), six lines, divided by the seven known products. Now, breaking into prime factors this result 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5. No 5 can go to the cage “2:” so both 5’s must go to the cage “16+” (numbers in blue) with a 6 (which gives the difference to 16). In these conditions the numbers inside “2:” must a have a product of 2700 / (5 x 5 x 6) = 2700 / 150 = 18, that is, “2:” = [1,3,6], [2,3,3,]. Obviously the only solution for “2:” is [1,3,6].

Next: The product of the cells c4 and d4 (looking to the three bottom rows) is:

Now we are going to define the other three corners since we can calculate the product of the corners. We must remember (see my post “Why in a 4x4 numbers in corners are those in inner area (2)” in the section “Solving strategies and tips”) that the product of the corners, named Xc, multiplied by the product of the numbers in the border, named Xb, equals (in a 6x6) to (6!) ^ 4 = 720 x 720 x 720 x 720. The reason for this is that if we multiply all numbers in the exterior rows, 1 and 6, and all numbers in the exterior columns, a and f, this total product is equal to 720 ^ 4, having multiplied all numbers in the border but including twice the corners so we have the product of the corners by the product of the border itself, Xc by Xb. Then, in our puzzle:

Xc = (720 x 720 x 720 x 720) / (96 x 900 x 288 x 540) = 20.

The product of the corners is 20 and being a6 = 5 we are left with two possibilities: (a1 =1; f1 = 4; f6 = 1) or (a1 = 2; f1 = 1; f6 = 2). If f1 = 4 we are dealing with the combination [3,3,4,5,5] for the cage “900x”(the position of the numbers is shown in red colour in the above graphic) and it can be quickly observed that now it would not be possible to place a 2 in column f (f4 or f5). As a consequence (in green): a1 = 2, f1 = 1, f6 = 2. The corners are now defined, they “inform” the full solution, and we can affirm that the puzzle is solved.

A final comment: Though it is not really necessary to calculate the solution for the cage “12x” (in this process it will be derived soon and straight forward) it’s interesting to underline now that this cage must be [2,2,3] for three reasons:

a) “96x” contains a 1 (and only once, due to f1 = 1 and d2 = 1, so [1,1,4,4,6] would never be possible); no more 1’s can go in any other place of the three upmost rows (or, inversely, if “12x” is [1,2,6] or [1,3,4] >>> b3 = 1 and there is no place for a 1 in the other positions of the cage “96x” but any of the possible combinations for this cage require a 1).

b) Since we have two 2’s in the corners, a total of four 2’s must go to the “inner area” of the puzzle (this is explained in the thread “Why in a 4x4 numbers in corners are those in inner area (1)”, in the same section, where it is shown that, in a 6x6, the 16 numbers in the “inner area”, that is the 4x4 inner box, are twice the set 1, 2, …, 6 plus the four corners) so both 2’s must go to the cage “12x”.

I had no "clue" regarding the addition rule or the multiplication rule but your explanation on the method of determining the corners makes sense to me now......not sure I'll remember how to do it the next time but I have made a copy for future reference....your posts in the past re: mod and bit/wise puzzles have been a great help so there is reason to believe that this one will also be of help.......thank you.

jomapil

Posted on:Mon Apr 16, 2012 10:26 am

Posts: 248Location: Lisbon, PortugalJoined: Sun Sep 18, 2011 3:40 pm

Re: The "power" of the corners: full analysis 6 x 6 2012-Apr

Another smart solution!

I solved this puzzle by " trial and error ". I must dedicate more to the analysis of the diagrams.I use few times the rule of multiplication. With the diagram 8x8 medium of 2012APR14 and little more.I saw that the " rule " of the corners must be used in these ( 6x6 ) type of puzzles.

I had no "clue" regarding the addition rule or the multiplication rule but your explanation on the method of determining the corners makes sense to me now......not sure I'll remember how to do it the next time but I have made a copy for future reference....your posts in the past re: mod and bit/wise puzzles have been a great help so there is reason to believe that this one will also be of help.......thank you.

I am really fascinated by the beauty of this puzzle and that’s what I try to transmit. Of course this method can only be applied in very particular cases, i.e., when we have products in the border, etc., though, in general, the “multiplication rule” can be applied whenever there are many products distributed over the puzzle.

jomapil wrote:

Another smart solution!

I solved this puzzle by " trial and error ". I must dedicate more to the analysis of the diagrams.I use few times the rule of multiplication. With the diagram 8x8 medium of 2012APR14 and little more.I saw that the " rule " of the corners must be used in these ( 6x6 ) type of puzzles.

Thank you again, Clm.

Sometimes before entering in the “trial and error” process we can eliminate possibilities. Now that we have seen the interest of the corners, in order to appreciate the beauty of this calcudoku, we may enjoy a little more its solution with an even more concise and faster way (two graphics, very few calculations):

Next we proceed with the corners: The 5 must go to a6 (it is not a submultiple of 96 or 288 and it is not possible in f1 because there must be two 5’s in “900x”); and reasoning as in the previous post, we observe (numbers in red colour in the above graphic) that (a1 = 1; f1 = 4; f6 = 1) is not possible since, being “900x” = [3,3,4,5,5] (unique in this hypothesis), there would not be place for a 2 in column f. Then, a1 = 2, f1 = 1, f6 = 2 (in green). This has two consequences (due also to d2 = 1):

a) Cage “96x”: Since it contains a 1 (now obviously [1,1,4,4,6] is not valid for this cage) >>> “12x” = [2,2,3] (no more than three 1’s in the three upmost rows), so b2 = 3, b3 = 2, c2 = 2.

b) c3 = 4 and c4 = 1 >>> d3 = 5 and d4 = 4.

And that’s all: The puzzle is solved (f1 = 1 >>> “900x” = [1,5,5,6,6]). Next graphic. We complete first the righmost part (gray and indigo colours) and later the leftmost part (orange and light blue colours):

What to do when 3 cages are multiplication and 4th cage is addition around the perimeter?

jotempe

Posted on:Mon Apr 16, 2012 8:28 pm

Posts: 31Joined: Mon Mar 05, 2012 10:45 am

Re: The "power" of the corners: full analysis 6 x 6 2012-Apr

The locations of all 5's n this puzzle can be quickly determined without multiplicaton rule - just notice, that multiplications have 4 factors of 5 among them, so two fives must be in the sum - and as there is only one sum in the diagram, their locations are determined. Starting from that you can quickly put all fives in their correct places. I don't remember the exact logic now, but I was able to solve the puzzle without trial and error and without multplication rule, through your way is probably better than the one I used, because it took me a while to determine the complete solution. And I agree, this one was really nice.

picklepep

Posted on:Tue Apr 17, 2012 2:59 am

Posts: 98Joined: Thu May 12, 2011 10:48 pm

Re: The "power" of the corners: full analysis 6 x 6 2012-Apr

beaker wrote:

What to do when 3 cages are multiplication and 4th cage is addition around the perimeter?

I tend to bisect the puzzle to try and find the inner square first.

jomapil

Posted on:Tue Apr 17, 2012 10:26 am

Posts: 248Location: Lisbon, PortugalJoined: Sun Sep 18, 2011 3:40 pm

Re: The "power" of the corners: full analysis 6 x 6 2012-Apr

" Elementary, my dear Watson !"After Clm, Jotempe and other colleagues the puzzles, even the very difficult, become easy.

Jotempe : I didn't see that the 5's were immediately known their position.

Is it possible, now and then, other people present some hints ( it's not necessary the complete solution ) about some puzzles? It would be excellent for the formation of the beginners ( and me ).

What to do when 3 cages are multiplication and 4th cage is addition around the perimeter?

I think that would depend on the particular puzzle, each case is different of course, the idea is having a good set of tools (as many as possible) to apply in different situations. If the “products” can be converted in sums or, inversely, the “sums” in products, we could use the "addition rule" or the "multiplication rule", otherwise we could go to complex equations.

But let’s suppose a case like the one in the next graphic, where the cage “900x” has been modified to “23+”, really it does not make any big difference, in this case we have a sum in the border, but once we have filled the “easy” numbers (i.e., positioning the six 5’s as in the jotempe’s explanation, identically valid in this case, and then the rest of numbers shown in blue), and once we have determined also that “80x” cann’t be [2,2,4,5] so it must be [1,4,4,5], we observe that the easiest way to solution the puzzle (this specific puzzle) would be to calculate the sum for the cage “288x” which must be 16 (the difference to 63, which is the total of the three rightmost columns) so “288x” = [2,3,3,4,4], unique among the four possibilities with five cells and that L-shape (previous post, once supressed the [2,2,2,6,6] that would never be valid with the L-shape). Now the puzzle is quicky finished.